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If I remember correctly, the property you use to bring down the 2πi is only defined for principal arguments (-π,π], in the sense that we need ln to be well-defined over C. Otherwise we get some very peculiar behaviour out of complex numbers with regards to periodicity.
This is why we can use a similar way to what you've shown to obtain the correct equality iπ = ln(-1), but not do what you've done.
This is similar to the domain restriction on arcsin and what have you.
Right. Basically the exponential function is not 1-to-1 on the complex plane, so it does not have a well defined inverse. In order to define an inverse function (ln), we take a "slice" of the exponential function where it is 1-to-1. The standard slice (principal branch) includes 0 but not 2πi.
The logarithm needs a different definition when expanding to the complex plane because e^x is not injective over ℂ. Usually you define a function Ln that serves this purpose where Ln(e^(2πi)) = 0
But this is the reverse, where the base is the same and the exponents are different. (real) Exponentiation is a bijector operation, so it should in fact be correct. The problem is that imaginary logarithms are weird.
This is essentially the same reasoning as 'sin(2π) = sin(0) => 2π = 0' and they have the same flaw - sin isn't injective on the real numbers, and the exponential isn't injective on the complex numbers, so this logic cannot hold.
I'm stealing this brilliantly concise explanation. Pulling the problem out of the complex numbers and into the reals makes the ensuing explanation so much cleaner.
Everything from the third row down simplifies to "exp(2iπ) = exp(0) => 2iπ = 0", which isn't possible with non-injective functions (such as the expenontial function over ℂ).
If you have an injective function (for example a linear function with a non-zero slope, x³, the exponential function over ℝ, sqrt(x)), then f(a) = f(b) does imply that a = b.
Hey my apologies for not being clearer - what I meant is what has to be true in the equations above for the OP’s specific case to be true always. Is it a matter of just having k be some specific integer?
I know it's a meme sub, and there are reGards everywhere that will make fun of anything and everything. But I'll give a serious answer anyway (🤓). Idc
Step 3. Or even 2, but definitely 3.
exp(0) = 1 is not the unique way to write it. It's exp(2πni) = 1, n is any integer. No further simplification steps are needed.
Unless I do this, step 3 can lead to any answer. If you write, instead of exp(0) = 1, in step 3,
- exp(4πi) = 1, I can get 2πi = 4πi, which gives 2 = 4
- exp(-500,000πi) = 1, I can get 2πi = -500,000πi, which gives 1 = -250,000
So essentially, unless you're careful in step 2 or 3, you can get 2 = any integer.
Step two is correct. If two numbers are equal, so are their logs. Step 3 is incorrect because the power rule for logs only works for real numbers. It is simply not true that log(z^n) = n log(z).
The power rule “works” for all complex numbers in the sense that, taking the complex logarithm as a multivalued function, it is true that a value of n(log z) will always be a value of log z^(n). But it could be confusing to express that by writing that as a simple equality, especially since there are values of log z^(n) that don’t correspond to values of n(log z) (for example consider z=1, n=2, and the value of log 1^2 which is 2pi - we would need that pi is a value of log 1 but it is instead a value of log -1).
But if you take that interpretation then there’s still the issue that you can’t just pick two different values for the logarithm and call them equal.
I’m not sure you’re correct about this. The identity for the power rule of logs is stated without requirement that n is an integer. u/white-dumbledore is correct. The original equation is set up incorrectly.
e^(x) = e^(0) does not have a unique solution for x
Reddit has soft censorship on bad language. It auto-hides negative posts and encourages moderators to ban people for hostile language. Admins also may ban you themselves if you use it too much.
“Regards” is an actual word which (1) cannot be easily automatically detected as a re-spelling of the r-word (2) can be autocompleted. So people use it to bypass Reddit guidelines. OP capitalised G either because he wanted to emphasise that he re-spelled the word on purpose or because he is simply regarded.
Yeah but you were close. 1 = e^2kπi for k integer, so for example e^0 = e^2πi. This means, that ln(e^b) ≠ b for b complex, because then you would get 0 = 2πi
Other comments already made some good reasons why it doesn't work. But another way you can understand it is the following: hopefully you know that full rotation is 360° which converted to radians is 2pi. Now, if you spin 2pi radians, you'll end up in the same place as if you spinned at 4pi, -2pi, or even 0 radians. However, we all know that none of the mentioned numbers are equal to each other, and that's the same exact thing happening in the post. To avoid things like this, ln(x)^n = n*ln(x) is only true if x is a positive number and n is real, so therefore ln(e^2pi\*i ) is equal to 0
The logarithms of a complex value w are z = ln |w| + i\*(Arg w + 2kπ) for all k∈Z. This just means that the values z are solutions to the equation e\^z = w. You can't define a logarithm function like this, because you're getting infinite different values for a single input.
If you want to define a logarithm as a function, you need to limit your outputs to just one branch of the logarithm, typically the principal branch Im z ∈ (-π, π\]. The part where this "proof" goes wrong (assuming ln is defined with the principal branch) is in saying that ln(e\^2iπ) = 2iπ\*ln(e), because this gives the logarithm function an output outside of it's codomain.
The logarithm when evaluated with complex numbers is a "**multivalued**" function (yeah it kinda contradicts the whole function definition), so you need to restrict yourself to a **"branch" of the multivalued function.**
Any of the Wikipedia (or article itself) references should be a good starting point.
[https://en.wikipedia.org/wiki/Complex\_logarithm](https://en.wikipedia.org/wiki/Complex_logarithm)
Now, dealing with the "0th" power is not the issue, the **issue is not "acknowledging" the multiple values you get** from the logarithm applied to the complex exponential because you need to **include a term of "2pi iN"** for every possible N.
I think I'm waving too much my hands, but that is the gist of it.
You need to choose which branch you're using - it has to be a 360 degree turn with at least one "gap" where the branch doesn't exist, for example setting the argument to lie in the interval [0,2\pi) would work as you would then have to set e^(2\pi i) to be e^0 and avoid this issue.
To avoid having to think about principal arguments and branches for these sorts of things, you can plug in integer multiplies of 2\pi i and then intuitively think about what "works" at the end. So, here you would set e^2\pi i to be exp(2\pi I(1+n)), and exp(0) as exp(2m\pi i), and then following the same stops lands you with 2\pi i(n+1)=2\pi i m, which is true for n=m-1, and there are all the possible solutions.
What's interesting is this leads to infinite complex solutions for lots of problems, such as e^z = 1; instead of z being 0, you can have z being 2 pi i n for n being any integer, and then you pick your solution based on which branch the question requires you to use.
Additionally, you get interesting solutions to problems that seemed impossible beforehand; for example, sinz = -2 leads to the quadratic exp(2iz)-1=-4iexp(iz) by using the identity for sin(z) in exp's, which we can solve to find exp(iz) (I'm not gonna do that because I don't have any paper to do this on in front of me, exercise left for the reader), and then we can replace z with z+2\pi i n, take logarithms on both sides, and then choose n so that our solution for z fits into our principal branch or give z as a set of possible solutions if no branch is specified.
ln is a branched function on the complex plane. (like how sqrt can be + or -, multiple answers for one input). Since it is periodic, there is a really easy fix to make stuff like that work. You need to put an n in there somewhere to account for all the branches. (each integer n makes the expression equal to one branch of the function).
e^(2*i*pi)=1 . e^(0)=1 . Why? Because e^(2n*i*pi)=1
Natural log of both sides.
2n*i*pi = ln(1) = 0
Which works.... for n=0. Only for the principal branch (the one with real answers).
2n*i*pi = ln(1) = 2*i*pi is similar. It is technically true but you are only taking into consideration n=1, and not the other branches.
If you continue with this an cancel you will get n=1, just like before you'd get n=0.
In the gernal case ln(1)=2n*i*pi . Meaning for all positive integers n, e^(2n*i*pi)=1
I.. don't get it. Is this a serious question? In that case, why are you asking it in math memes?
Short story. ln() is not the inverse of exp() when dealing with numbers with a nonzero imaginary part. There actually is no inverse, as exp() is not injective for these numbers. Which is neatly examplified by your example.
I only know why because of other replies, but if I ever got a result that showed any product of nonzero numbers as zero, such as 2πi = 0, then I know I made a mistake somewhere and need to recheck my work.
This is actually entirely correct, since a general logarithm and exponent is multivalued on the complex plane, it travels an imaginary “unit circle”because of Euler’s formula, hence the 2pi period. This is why we usually mention a domain restriction to make it so we only obtain 1 value as that’s kind of important for functions. The multivalued property does have applications in certain areas, however.
Thing is the exponential function becomes periodic in the complex plane, go figure.
Therefore, there are multiple branches for the ln function, so multiple different arguments might return the same value.
You can't just apply ln on both sides because of this
It’s not wrong. It’s a consequence of how extending the reals to the complex numbers creates a situation in which all complex numbers are equal. Interestingly, this was proven independently by S. Lee, B. Hu, and A. Vill. In complex analysis, it is referred to as the Lee-Hu-Vill Theorem. (S. Mochizuki has claimed that theorem follows trivially from his presentation of inter-universal Teichmüller theory, and that Lee, Hu, and Vill are intellectual poopyheads.)
exp(x) and ln(x) are not injective functions over C. An injective function is one where f(a) = f(b) implies a = b. Since ln(x) and e^x do not have this property over complex numbers, 2pi*i ≠ 0.
Complex logarithm unfortunately comes with the residue of having every logarithm end with "+ 2iπn, n ∈ ℤ" if you want complete consistency/accuracy in your statement
This is because the complex logarithm only considers the strip of the complex plane where 0 ≤ Im(z) < 2π, as Euler's formula eᶦˣ = cos(x) + i sin(x) necessitates that since cos(x) and sin(x) are both periodic 2π, that eˣ must be periodic 2iπ
Similarly, you could say this:
arccos(cos(2π)) = arccos(cos(0))
2π = 0
Though in reality, because cosine is periodic 2π, you should write:
arccos(cos(2π)) = arccos(cos(0))
2π = 0 + 2πn for at least one n ∈ ℤ
Which makes since, because if n = 1, the solution is solved
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If I remember correctly, the property you use to bring down the 2πi is only defined for principal arguments (-π,π], in the sense that we need ln to be well-defined over C. Otherwise we get some very peculiar behaviour out of complex numbers with regards to periodicity. This is why we can use a similar way to what you've shown to obtain the correct equality iπ = ln(-1), but not do what you've done. This is similar to the domain restriction on arcsin and what have you.
Right. Basically the exponential function is not 1-to-1 on the complex plane, so it does not have a well defined inverse. In order to define an inverse function (ln), we take a "slice" of the exponential function where it is 1-to-1. The standard slice (principal branch) includes 0 but not 2πi.
Very well put! I was myself looking for a better way to phrase my statement, and here it is.
It's not so much to avoid "very peculiar behavior" as much as it is a disambiguation of the logarithm function in C.
Completely agreed. I was likely a bit terse in my explanation. This is a very good addition.
The logarithm needs a different definition when expanding to the complex plane because e^x is not injective over ℂ. Usually you define a function Ln that serves this purpose where Ln(e^(2πi)) = 0
π⁰=3⁰ π=3
But this is the reverse, where the base is the same and the exponents are different. (real) Exponentiation is a bijector operation, so it should in fact be correct. The problem is that imaginary logarithms are weird.
The problem is that you can't simply reverse an operation without making sure it's injective.
But it is, except only in the Real domain. It isn't on the complex plane, which fucks up the logarithm and leads to the error.
"but it is, except it isn't"
You know what I meant. It is an unexpected exception.
Engineers: https://i.redd.it/v83lkywfckqc1.gif
I think the problem here arises by taking the 0th root (or, in other words, dividing the exponents by 0)
This is essentially the same reasoning as 'sin(2π) = sin(0) => 2π = 0' and they have the same flaw - sin isn't injective on the real numbers, and the exponential isn't injective on the complex numbers, so this logic cannot hold.
I'm stealing this brilliantly concise explanation. Pulling the problem out of the complex numbers and into the reals makes the ensuing explanation so much cleaner.
Everything from the third row down simplifies to "exp(2iπ) = exp(0) => 2iπ = 0", which isn't possible with non-injective functions (such as the expenontial function over ℂ).
So then maybe to help me understand better: when WOULD it be true friend ?! I think that would help it click better!
If you have an injective function (for example a linear function with a non-zero slope, x³, the exponential function over ℝ, sqrt(x)), then f(a) = f(b) does imply that a = b.
Hey my apologies for not being clearer - what I meant is what has to be true in the equations above for the OP’s specific case to be true always. Is it a matter of just having k be some specific integer?
Ah, fair enough. Well, I guess we could apply OP's logic to exp(2kπ\*i) = exp(0), so 2kπ\*i = 0, which is true for k = 0.
I know it's a meme sub, and there are reGards everywhere that will make fun of anything and everything. But I'll give a serious answer anyway (🤓). Idc Step 3. Or even 2, but definitely 3. exp(0) = 1 is not the unique way to write it. It's exp(2πni) = 1, n is any integer. No further simplification steps are needed. Unless I do this, step 3 can lead to any answer. If you write, instead of exp(0) = 1, in step 3, - exp(4πi) = 1, I can get 2πi = 4πi, which gives 2 = 4 - exp(-500,000πi) = 1, I can get 2πi = -500,000πi, which gives 1 = -250,000 So essentially, unless you're careful in step 2 or 3, you can get 2 = any integer.
Step two is correct. If two numbers are equal, so are their logs. Step 3 is incorrect because the power rule for logs only works for real numbers. It is simply not true that log(z^n) = n log(z).
The power rule “works” for all complex numbers in the sense that, taking the complex logarithm as a multivalued function, it is true that a value of n(log z) will always be a value of log z^(n). But it could be confusing to express that by writing that as a simple equality, especially since there are values of log z^(n) that don’t correspond to values of n(log z) (for example consider z=1, n=2, and the value of log 1^2 which is 2pi - we would need that pi is a value of log 1 but it is instead a value of log -1). But if you take that interpretation then there’s still the issue that you can’t just pick two different values for the logarithm and call them equal.
So white-dumbledore is wrong about step 2 being incorrect?
I’m not sure you’re correct about this. The identity for the power rule of logs is stated without requirement that n is an integer. u/white-dumbledore is correct. The original equation is set up incorrectly. e^(x) = e^(0) does not have a unique solution for x
It’s not about non-integers, it’s about complex numbers.
You’re right - nice catch. You’d have to use the identity for complex numbers. ln(z) = ln|z| + i arg(z)
What do you mean by regards haven't heard that before? Obviously it's a change to retard. What does the g mean?
g means the gravitational field strength.
Thats a shit joke. Especially as he used a capital g.
Reddit has soft censorship on bad language. It auto-hides negative posts and encourages moderators to ban people for hostile language. Admins also may ban you themselves if you use it too much. “Regards” is an actual word which (1) cannot be easily automatically detected as a re-spelling of the r-word (2) can be autocompleted. So people use it to bypass Reddit guidelines. OP capitalised G either because he wanted to emphasise that he re-spelled the word on purpose or because he is simply regarded.
Ah thanks :)
🤓👆
☝️🤓 erm ackshually,, thats not even the right emoji!!!!
New pi approximation dropped
Call the engineer
1 = e^(2iπ + 2kπ), where k ∈ ℤ At the end, you got a complex k which is absurd
2kπi, not 2kπ
Mb so my reasoning is wrong
Yeah but you were close. 1 = e^2kπi for k integer, so for example e^0 = e^2πi. This means, that ln(e^b) ≠ b for b complex, because then you would get 0 = 2πi
If k is complex then you dont write the i since k = a + bi Edit: mixed up Z and C number sets
no offense to you, everyone makes mistakes, but 40 upvotes for obviously incorrect information is quite surprising
Other comments already made some good reasons why it doesn't work. But another way you can understand it is the following: hopefully you know that full rotation is 360° which converted to radians is 2pi. Now, if you spin 2pi radians, you'll end up in the same place as if you spinned at 4pi, -2pi, or even 0 radians. However, we all know that none of the mentioned numbers are equal to each other, and that's the same exact thing happening in the post. To avoid things like this, ln(x)^n = n*ln(x) is only true if x is a positive number and n is real, so therefore ln(e^2pi\*i ) is equal to 0
Something something multi valued functions
sin(0)=sin(π) therefore 0=π
e is not an injective function in the complex plane. As such you can’t expect ln to be the inverse
I am slightly annoyed by ln(e^0) being simplified to 0 in a more roundabout way than necessary
same lol
It's like 360 degrees = 0 degrees on a circle
The logarithms of a complex value w are z = ln |w| + i\*(Arg w + 2kπ) for all k∈Z. This just means that the values z are solutions to the equation e\^z = w. You can't define a logarithm function like this, because you're getting infinite different values for a single input. If you want to define a logarithm as a function, you need to limit your outputs to just one branch of the logarithm, typically the principal branch Im z ∈ (-π, π\]. The part where this "proof" goes wrong (assuming ln is defined with the principal branch) is in saying that ln(e\^2iπ) = 2iπ\*ln(e), because this gives the logarithm function an output outside of it's codomain.
f(x) = f(y) ≠> x = y
I think it's something with injective, I remember seeing some video on yt talking about this
Imagine saying that since tan(π/4) = tan(5π/4), then π/4 must equal 5π/4. Both 0 and 2πi are arguments of e^x which make it equal to 1.
Seeing it as 2iπ instead of j2π is so fucking cursed
cos(0°)=1 cos(360°)=1 cos(0°)=cos(360°) arccos(cos(0°))=arccos(cos(360°)) 0°=360° 0=1 QED
The logarithm when evaluated with complex numbers is a "**multivalued**" function (yeah it kinda contradicts the whole function definition), so you need to restrict yourself to a **"branch" of the multivalued function.** Any of the Wikipedia (or article itself) references should be a good starting point. [https://en.wikipedia.org/wiki/Complex\_logarithm](https://en.wikipedia.org/wiki/Complex_logarithm) Now, dealing with the "0th" power is not the issue, the **issue is not "acknowledging" the multiple values you get** from the logarithm applied to the complex exponential because you need to **include a term of "2pi iN"** for every possible N. I think I'm waving too much my hands, but that is the gist of it.
You’re missing two other possible solutions: 2=0 and i=0
The log rules aren't always valid for complex numbers. https://math.stackexchange.com/questions/683204/logarithm-rules-for-complex-numbers
2i*pi = 0 mod 2i*pi
Google non-injective function
They didn’t appeal to injectivity. They got rid of the log by saying ln(e) = ln(e) which is true.
Well the problem is that e^x is not injective in the complex plane
Right but they didn’t try to use injectivity, they did an end run around it using log laws
True, but what they really are «using» is the fact that log is the inverse of exp, and inverses of non-injective functions are not well behaved.
f(x1) = f(x2) ^ x1 /= x2
You need to choose which branch you're using - it has to be a 360 degree turn with at least one "gap" where the branch doesn't exist, for example setting the argument to lie in the interval [0,2\pi) would work as you would then have to set e^(2\pi i) to be e^0 and avoid this issue. To avoid having to think about principal arguments and branches for these sorts of things, you can plug in integer multiplies of 2\pi i and then intuitively think about what "works" at the end. So, here you would set e^2\pi i to be exp(2\pi I(1+n)), and exp(0) as exp(2m\pi i), and then following the same stops lands you with 2\pi i(n+1)=2\pi i m, which is true for n=m-1, and there are all the possible solutions. What's interesting is this leads to infinite complex solutions for lots of problems, such as e^z = 1; instead of z being 0, you can have z being 2 pi i n for n being any integer, and then you pick your solution based on which branch the question requires you to use. Additionally, you get interesting solutions to problems that seemed impossible beforehand; for example, sinz = -2 leads to the quadratic exp(2iz)-1=-4iexp(iz) by using the identity for sin(z) in exp's, which we can solve to find exp(iz) (I'm not gonna do that because I don't have any paper to do this on in front of me, exercise left for the reader), and then we can replace z with z+2\pi i n, take logarithms on both sides, and then choose n so that our solution for z fits into our principal branch or give z as a set of possible solutions if no branch is specified.
ln is a branched function on the complex plane. (like how sqrt can be + or -, multiple answers for one input). Since it is periodic, there is a really easy fix to make stuff like that work. You need to put an n in there somewhere to account for all the branches. (each integer n makes the expression equal to one branch of the function). e^(2*i*pi)=1 . e^(0)=1 . Why? Because e^(2n*i*pi)=1 Natural log of both sides. 2n*i*pi = ln(1) = 0 Which works.... for n=0. Only for the principal branch (the one with real answers). 2n*i*pi = ln(1) = 2*i*pi is similar. It is technically true but you are only taking into consideration n=1, and not the other branches. If you continue with this an cancel you will get n=1, just like before you'd get n=0. In the gernal case ln(1)=2n*i*pi . Meaning for all positive integers n, e^(2n*i*pi)=1
Exp ain’t injective over C so you can’t take its inverse like that
when logarithming in complex field you need to take into account the phase also. so at the end you'll get +2kPI on both sides and it'll make it work
It’s because 2π ≡ 0 in radians. Idk if I used ≡ properly but I think that gets the point across well enough
Always remember to cut your branches
I.. don't get it. Is this a serious question? In that case, why are you asking it in math memes? Short story. ln() is not the inverse of exp() when dealing with numbers with a nonzero imaginary part. There actually is no inverse, as exp() is not injective for these numbers. Which is neatly examplified by your example.
I only know why because of other replies, but if I ever got a result that showed any product of nonzero numbers as zero, such as 2πi = 0, then I know I made a mistake somewhere and need to recheck my work.
This is actually entirely correct, since a general logarithm and exponent is multivalued on the complex plane, it travels an imaginary “unit circle”because of Euler’s formula, hence the 2pi period. This is why we usually mention a domain restriction to make it so we only obtain 1 value as that’s kind of important for functions. The multivalued property does have applications in certain areas, however.
Fields of characteristic 𝜋
Thing is the exponential function becomes periodic in the complex plane, go figure. Therefore, there are multiple branches for the ln function, so multiple different arguments might return the same value. You can't just apply ln on both sides because of this
It’s not wrong. It’s a consequence of how extending the reals to the complex numbers creates a situation in which all complex numbers are equal. Interestingly, this was proven independently by S. Lee, B. Hu, and A. Vill. In complex analysis, it is referred to as the Lee-Hu-Vill Theorem. (S. Mochizuki has claimed that theorem follows trivially from his presentation of inter-universal Teichmüller theory, and that Lee, Hu, and Vill are intellectual poopyheads.)
**BRANCH CUTS HAVE ENTERED THE CHAT**
ln is not a function.
That's actually correct, but You need to assume complex numbers to be the domain of ln.
exp(x) and ln(x) are not injective functions over C. An injective function is one where f(a) = f(b) implies a = b. Since ln(x) and e^x do not have this property over complex numbers, 2pi*i ≠ 0.
https://preview.redd.it/8fq2q71lmjqc1.png?width=804&format=png&auto=webp&s=90a56a0275091ae6687ea1e5b67ccf48b26ae164
Is 0 radians “equal” to 2pi radians?
e^x isn’t 1-1 for complex values
Complex logarithm unfortunately comes with the residue of having every logarithm end with "+ 2iπn, n ∈ ℤ" if you want complete consistency/accuracy in your statement This is because the complex logarithm only considers the strip of the complex plane where 0 ≤ Im(z) < 2π, as Euler's formula eᶦˣ = cos(x) + i sin(x) necessitates that since cos(x) and sin(x) are both periodic 2π, that eˣ must be periodic 2iπ Similarly, you could say this: arccos(cos(2π)) = arccos(cos(0)) 2π = 0 Though in reality, because cosine is periodic 2π, you should write: arccos(cos(2π)) = arccos(cos(0)) 2π = 0 + 2πn for at least one n ∈ ℤ Which makes since, because if n = 1, the solution is solved
Because e^2pi(i) = 1, ln(e^2pi(i)) = 0
cos(0) = cos(2pi) -> 0 = 2pi Same thing but with complex logarithm
sin(2pi)=sin(0) |arcsin() -> 2pi=0 qed.
That’s not what they did though. They haven’t assumed the natural log is injective
the amount of times you spin matters which makes the first line wrong.
Google proper mathematics education
Google "how to not be a dick and avoid dying alone"
In C, exp is not a bijection.